The question is incomplete. The complete question is :
A wire 0.6 m long and with mass m = 11 g is positioned horizontally near the earth's surface and perpendicular to a horizontal magnetic field of magnitude B = 0.4 T. What current I must flow through the wire in order that the wire accelerate neither upwards nor downwards? The magnetic field is directed into the page.
Solution :
Given :
Length of the wire, L = 0.6 m
Mass of the wire length, m = 11 g
= [tex]11 \times 10^{-3}[/tex] kg
Magnetic field , B = 0.4 T
Know we know that :
ILB = mg
or [tex]$I=\frac{mg}{BL}$[/tex]
[tex]$I= \frac{(11 \times 10^{-3})(9.81)}{(0.4)(0.6)}$[/tex]
[tex]I=0.44963\ A[/tex]
[tex]I = 449.63 \ mA[/tex]
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
In Young's double slit experiment, 402 nm light gives a fourth-order bright fringe at a certain location on a flat screen. What is the longest wavelength of visible light that would produce a dark fringe at the same location? Assume that the range of visible wavelengths extends from 380 to 750 nm.
Answer:
λ₂ = 357.3 nm
Explanation:
The expression for double-slit interference is
d sin θ = m λ constructive interference
d sin θ = (m + ½) λ destructive interference.
The initial data corresponds to a constructive interference, they indicate that we are in the fourth order (m = 4), let's look for the separation of the slits
d sin θ = m λ₁
now ask for destructive interference for m = 4
d sin θ = (m + ½) λ₂
we match these two expressions
m λ₁ = (m + ½) λ₂
λ₂ = ( m / m + ½) λλ₁
let's calculate
λ₂ =[tex]\frac{4}{(4.000 +0.5) \ 401}[/tex]
λ₂ = 357.3 nm
Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
A metal blade of length L = 300 cm spins at a constant rate of 17 rad/s about an axis that is perpendicular to the blade and through its center. A uniform magnetic field B = 4.0 mT is perpendicular to the plane of rotation. What is the magnitude of the potential difference (in V) between the center of the blade and either of its ends?
We are being given that:
The length of a metal blade = 300 cmThe angular velocity at which the metal blade is rotating about its axis is ω = 17 rad/sThe magnetic field (B) = 4.0 mTA pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.
From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).
The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:
[tex]\mathsf{E = B\times l\times v}[/tex]
where;
B = magnetic fieldl = lengthv = relative speedFrom the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.
Then, the velocity (v) = ωx
The potential difference across it can now be expressed as:
[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]
For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.
∴
[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]
[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]
[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]
Recall that,
magnetic field (B) = 4 mT = 4 × 10⁻³ TLength L = 300 cm = 3mangular velocity (ω) = 17 rad/s[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]
[tex]\mathsf{E = 19.13 mV}[/tex]
Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.
Learn more about Faraday's law of induction here:
https://brainly.com/question/13369951?referrer=searchResults
Parallel Wires: Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other?
Answer:
Explanation:
Given force between 2 currents carrying
wires = F₀
Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L
where I₁=Current in wire 1
I₂= Current in wire 2
L= Length of the wire
when one current is doubled and the other is halved
I₁= 2 I₁
I₂= I₂/2
F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L
1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N
Answer:
D) 1003 N
Explanation:
Given the following data;
Mass of man = 85 kg
Acceleration of elevator = 2 m/s²
Acceleration due to gravity, g = 9.8 m/s²
To find the force exerted by the man on the floor;
Force = mg + ma
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
An ideal double slit interference experiment is performed with light of wavelength 640 nm. A bright spot is observed at the center of the resulting pattern as expected. For the 2n dark spot away from the center, it is known that light passing through the more distant slit travels the closer slit.
a) 480 nm
b) 600 nm
c) 720 nm
d) 840 nm
e) 960 nm
Answer:
960 nm
Explanation:
Given that:
wavelength = 640 nm
For the second (2nd) dark spot; the order of interference m = 1
Thus, the path length difference is expressed by the formula:
[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]
[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]
[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]
dsinθ = 960 nm
In the Bohr model of the hydrogen atom, an electron in the 3rd excited state moves at a speed of 2.43 105 m/s in a circular path of radius 4.76 10-10 m. What is the effective current associated with this orbiting electron
Answer:
Current =,charge / time
Charge = e = 1.6E-19 coulombs
t = T time for 1 revolution (period)
v = S / T = distance traveled in 1 revolution / time for 1 revolution
T = S / v = 2 pi * 4.76E-10 / 2.43E5 = 1.23E-14
I = Q / T = 1.6E-19 / 1.23E-14 = 1.30E-5
Serena wants to play a trick on her friend Marion. She adds salt, sugar, and vinegar into her glass of water when Marion is out of the room. Why does she know that Marion will drink the water?
đổi đơn vị
42 ft2/hr to cm2/s
Answer:
X = 10.8387 cm²/s
Explanation:
In this exercise, you're required to convert a value from one unit to another.
Converting 42 ft²/hr to cm²/s;
Conversion:
1 ft² = 929.03 cm²
42 ft² = X cm²
Cross-multiplying, we have;
X = 42 * 929.03
X = 39019.26 cm²
Next, we would divide by time in seconds.
1 hour = 3600 seconds
X = 39019.26/3600
X = 10.8387 cm²/s
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bottemvof the ramp
A
Explanation:
1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.
Explain why liquid particles at a high pressure would need more
energy to change to a gas than liquid particles at a low pressure.
Answer:
Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.
Unit of speed is a derived unit. Give reasons
Answer:
as it 8s based upon to fundamental units distance and Time
Which one of the following physical quantities has its S.I. unit m/s?
(i) Acceleration
(ii) Velocity
(iii) Force
(iv) Density
Answer:
velocity is the answer of this question.
Answer:
Velocity is the right answer ok
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
The given parameters include:
constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 mTo calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]
To learn more about projectile calculations please visit: https://brainly.com/question/14083704
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s
Answer:
9 Brainly hahaha ............huh
an aluminum atom has an atomic number of 13 and a mass number of 27,how many
a)protons
b) electrons
pls write the formula too
Element is
[tex]\boxed{\sf {}^{27}Al_{13}}[/tex]
Atomic number=13Mass number=27[tex]\\ \sf\longmapsto No\:of\:Protons=Atomic \:Number=13[/tex]
And[tex]\\ \sf\longmapsto No\:of\:Neutrons=Mass\:number-Atomic\:Number[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=27-13[/tex]
[tex]\\ \sf\longmapsto No\:of\:Neutrons=14[/tex]
And
[tex]\\ \sf\longmapsto No\:of\:electrons=No\:of\:Protons=13[/tex]
When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.
Solution :
Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.
And thus the bulb flashes for the moment as opposed to continuous glow.
Imagine a spaceship traveling at a constant speed through outer space. The length of the ship, as measured by a passenger aboard the ship, is 28.2 m. An observer on Earth, however, sees the ship as contracted by 16.1 cm along the direction of motion. What is the speed of the spaceship with respect to the Earth
[tex]3.20×10^7\:\text{m/s}[/tex]
Explanation:
Let
[tex]L = 28.2\:\text{m}[/tex]
[tex]L' = 28.2\:\text{m} - 0.161\:\text{m} = 28.039\:\text{m}[/tex]
The Lorentz length contraction formula is given by
[tex]L' = L\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)}[/tex]
where L is the length measured by the moving observer and L' is the length measured by the stationary Earth-based observer. We can rewrite the above equation as
[tex]\sqrt {1 - \left(\dfrac{v^2}{c^2}\right)} = \dfrac{L'}{L}[/tex]
Taking the square of the equation, we get
[tex]1 - \left(\dfrac{v^2}{c^2}\right) = \left(\dfrac{L'}{L}\right)^2[/tex]
or
[tex]1 - \left(\dfrac{L'}{L}\right)^2 = \left(\dfrac{v}{c}\right)^2[/tex]
Solving for v, we get
[tex]v = c\sqrt{1 - \left(\dfrac{L'}{L}\right)^2}[/tex]
[tex]\:\:\:\:=(3×10^8\:\text{m/s})\sqrt{1 - \left(\dfrac{28.039\:\text{m}}{28.2\:\text{m}}\right)^2}[/tex]
[tex]\:\:\:\:=3.20×10^7\:\text{m/s} = 0.107c[/tex]
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
which energy does a car travelling 30 m/ph as it slows have:
a). chemical energy
b). thermal energy
c). kinetic energy
please helpp
Answer:
c) kinetic energy
Explanation:
Answer: C) kinetic energy
Explanation:
state the laws of reflection
Answer:
Explanation:
The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.
All other properties of reflection flow from this one statement.
